What is the integral of (s^2-1)/((s^2+1)^2) ?

The integral of (s^2-1)/((s^2+1)^2) is -s/(1+s^2)+C

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integral of (s^2-1)/((s^2+1)^2)

\int \frac{s ^{2} - 1}{( s ^{2} + 1 ) ^{2}} d s

- \frac{s}{1 + s ^{2}} + C

Solution steps

\int \frac{s ^{2} - 1}{( s ^{2} + 1 ) ^{2}} d s

Take the partial fraction of

\frac{s ^{2} - 1}{( s ^{2} + 1 ) ^{2}} : \frac{1}{s ^{2} + 1} - \frac{2}{( s ^{2} + 1 ) ^{2}}

= \int \frac{1}{s ^{2} + 1} - \frac{2}{( s ^{2} + 1 ) ^{2}} d s

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{s ^{2} + 1} d s - \int \frac{2}{( s ^{2} + 1 ) ^{2}} d s

\int \frac{1}{s ^{2} + 1} d s = arctan (s)

\int \frac{2}{( s ^{2} + 1 ) ^{2}} d s = arctan (s) + \frac{s}{1 + s ^{2}}

= arctan (s) - (arctan (s) + \frac{s}{1 + s ^{2}})

Simplify

= - \frac{s}{1 + s ^{2}}

Add a constant to the solution

= - \frac{s}{1 + s ^{2}} + C

What is the integral of (s^2-1)/((s^2+1)^2) ?
The integral of (s^2-1)/((s^2+1)^2) is -s/(1+s^2)+C